/*
Define f(0)=1 and f(n) to be the number of ways to write n as a sum of powers of 2 where no power occurs more than twice. 

For example, f(10)=5 since there are five different ways to express 10:10 = 8+2 = 8+1+1 = 4+4+2 = 4+2+2+1+1 = 4+4+1+1

It can be shown that for every fraction p/q (p&gt;0, q&gt;0) there exists at least one integer n such that f(n)/f(n-1)=p/q.
For instance, the smallest n for which f(n)/f(n-1)=13/17 is 241.
The binary expansion of 241 is 11110001.
Reading this binary number from the most significant bit to the least significant bit there are 4 one's, 3 zeroes and 1 one. We shall call the string 4,3,1 the Shortened Binary Expansion of 241.
Find the Shortened Binary Expansion of the smallest n for which f(n)/f(n-1)=123456789/987654321.
Give your answer as comma separated integers, without any whitespaces.

Anser:
Time:
*/
package main

import (
	"fmt"
	"time"
)

func main() {
	tstart := time.Now()



	tend := time.Now()
	fmt.Println(tend.Sub(tstart))
}